Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $y = \dfrac{x - 7}{x + 7} \times \dfrac{x + 7}{x^2 - 13x + 42} $
Solution: First factor the quadratic. $y = \dfrac{x - 7}{x + 7} \times \dfrac{x + 7}{(x - 7)(x - 6)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (x - 7) \times (x + 7) } { (x + 7) \times (x - 7)(x - 6) } $ $y = \dfrac{ (x - 7)(x + 7)}{ (x + 7)(x - 7)(x - 6)} $ Notice that $(x + 7)$ and $(x - 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ \cancel{(x - 7)}(x + 7)}{ (x + 7)\cancel{(x - 7)}(x - 6)} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $y = \dfrac{ \cancel{(x - 7)}\cancel{(x + 7)}}{ \cancel{(x + 7)}\cancel{(x - 7)}(x - 6)} $ We are dividing by $x + 7$ , so $x + 7 \neq 0$ Therefore, $x \neq -7$ $y = \dfrac{1}{x - 6} ; \space x \neq 7 ; \space x \neq -7 $